Using the balanced equation shown below, what is the mass of C3H8 that must react in order to release 1.25×10^6 kJ of heat? ΔHrxn = –2219.9 kJ

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Using the balanced equation shown below, what is the mass of C3H8 that must react in order to release 1.25×10^6 kJ of heat? ΔHrxn = –2219.9 kJ

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Answer 1 · 2017-07-26T22:43:21

Well, apparently,

$m_{C_{3} H_{8}} = 24.8 kg$ $m_(C_3H_8) = "24.8 kg"$

Reactions tend to occur at constant pressure, so consequently, we write that

$q_(rxn) = DeltaH_(rxn)$

I assume your $DeltaH_(rxn)$ units are not correct and should be $"kJ/mol"$ ; otherwise, there would not be any point in knowing the mass of the reactant.

Define $DeltabarH_(rxn) = (DeltaH_(rxn))/n_(C_3H_8)$ , where $n_(C_3H_8)$ is mols of $"C"_3"H"_8$ . This means...

$n_(C_3H_8)DeltabarH_(rxn) = DeltaH_(rxn) = q_(rxn)$

$= n_(C_3H_8) xx (-"2219.9 kJ"/("mol C"_3"H"_8))$

$= -1.25 xx 10^6$ $"kJ"$

Therefore, this many mols of propane reacted:

$n_(C_3H_8) = -1.25 xx 10^6 cancel"kJ" xx ("1 mol C"_3"H"_8)/(-2219.9 cancel"kJ")$

$=$ $"563.09 mols"$

...wow, that's huge... Well, in that case...

$color(blue)(m_(C_3H_8)) = 563.09 cancel("mols C"_3"H"_8) xx ("44.1 g C"_3"H"_8)/cancel("1 mol C"_3"H"_8)$

$=$ $"24832.20 g C"_3"H"_8$

$= color(blue)("24.8 kg C"_3"H"_8)$

I would not want to be at this factory... they're combusting kilos of propane!

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Using the balanced equation shown below, what is the mass of C3H8 that must react in order to release 1.25×10^6 kJ of heat? ΔHrxn = –2219.9 kJ

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